Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $a \neq 0$. $k = \dfrac{a - 6}{6a - 48} \times \dfrac{3a^2 + 9a}{a^2 - 3a - 18} $
Solution: First factor the quadratic. $k = \dfrac{a - 6}{6a - 48} \times \dfrac{3a^2 + 9a}{(a - 6)(a + 3)} $ Then factor out any other terms. $k = \dfrac{a - 6}{6(a - 8)} \times \dfrac{3a(a + 3)}{(a - 6)(a + 3)} $ Then multiply the two numerators and multiply the two denominators. $k = \dfrac{ (a - 6) \times 3a(a + 3) } { 6(a - 8) \times (a - 6)(a + 3) } $ $k = \dfrac{ 3a(a - 6)(a + 3)}{ 6(a - 8)(a - 6)(a + 3)} $ Notice that $(a + 3)$ and $(a - 6)$ appear in both the numerator and denominator so we can cancel them. $k = \dfrac{ 3a\cancel{(a - 6)}(a + 3)}{ 6(a - 8)\cancel{(a - 6)}(a + 3)} $ We are dividing by $a - 6$ , so $a - 6 \neq 0$ Therefore, $a \neq 6$ $k = \dfrac{ 3a\cancel{(a - 6)}\cancel{(a + 3)}}{ 6(a - 8)\cancel{(a - 6)}\cancel{(a + 3)}} $ We are dividing by $a + 3$ , so $a + 3 \neq 0$ Therefore, $a \neq -3$ $k = \dfrac{3a}{6(a - 8)} $ $k = \dfrac{a}{2(a - 8)} ; \space a \neq 6 ; \space a \neq -3 $